Fundamentals of Electromagnetic Radiation

7 downloads 32 Views 86KB Size Report
ECE 144. Electromagnetic Fields and Waves. Bob York. Fundamentals of Electromagnetic Radiation. Hertzian Dipole. The most elementary source of radiation ...

ECE 144

Electromagnetic Fields and Waves

Bob York

Fundamentals of Electromagnetic Radiation Hertzian Dipole The most elementary source of radiation is the oscillating electric dipole, or Hertzian dipole. We can model this source as an infinitisimally thin filament of current of length dl in the zˆ direction, as shown below.

The current is assumed to be spatially constant over the length dl and time-harmonic. Mathematically the current density is  I δ(x )δ(y )ˆ z −dl/2 < z  < dl/2 (1) J(x , y , z  ) = 0 elsewhere where δ(x) is the Dirac delta function. This current produces the vector potential   dl/2 J(x , y , z  )e−jkR I zˆ e−jkr  µ µ A(x, y, z) = dV   dz 4π V  R 4π −dl/2 r µ e−jkr Idl zˆ 4π r Converting this to spherical coordinates and calculating the fields from =

H= gives

1 ∇×A µ



1 ∇×H jω

  I dl 2 1 1 e−jkr k sin θ + 4π jkr (jkr)2      1 Idl 2 1 ˆ sin θ 1 + 1 + 1 r cos θ + θ e−jkr E=− + ηk 2ˆ 4π (jkr)2 (jkr)3 jkr (jkr)2 (jkr)3

H = −φˆ


In the Far-Field of the antenna where kr  1, the terms in (3) that are proportional to 1/kr dominate, giving   I dl e−jkr H = Hφ φˆ = j k sin θ φˆ 4π r (4) Far-Field:   I dl e−jkr ˆ ˆ E = Eθ θ = j kη sin θ θ 4π r The radiation in the far-field behaves much like a plane wave, field intensity dropping off as 1/r. If  with the ∗ 1 using the complete field expressions we calculate the time-average Poynting vector P ave = 2 Re E × H (3), the same result is obtained as when the fields (4) are used. Thus the terms in (3) proportional to 1/r are considered as responsible for the average power transfer away from the antenna, while the other terms in (3) are interpreted as contributing to stored (reactive) energy in the immediate vicinity of the antenna.

Hertzian Dipole Parameters Using the far-field expressions (4) the time-averaged poynting vector is (I dl)2 2 sin2 θ 1  ∗ = rˆ P ave = Re E × H ηk 2 32π 2 r2 giving a total radiated power of  2π π 1 Pr = (P ave · rˆ) r 2 sin θ dθ dφ = Rr I 2 2 0 0 where Rr is defined as the radiation resistance, Rr = 80π The gain is found to be G(θ, φ) =


dl λ



2 (7)

4πr 2 P ave 3 = sin2 θ Pr 2


and so the directivity is D = Gmax = 1.5. The half-power beamwidth is found from (8) to be 90◦ . Equation (8) also indicates that the radiation pattern is constant with φ, and describes a torus as shown below.

If the dipole is made from a metal wire of radius a, the radiation efficiency is  −1

Rs λ2 ηr = 1 + where Rs = πfµ/σ 3 160π a dl


Magnetic Dipole A similar analysis of the short current loop pictured below reveals radiated fields nearly identical to (4), if the roles of electric and magnetic fields are interchanged. This duality is a result of the symmetry of Maxwell’s equations. As a result, we can think of the current loop as a magnetic dipole.

The far-field of the magnetic dipole is described by 2


120π IA e E = φˆ sin θ 2 λ r

Eφ H = −θˆ η


where A = πa2 is the area of the loop. The radiation resistance can then be found as Rr =

320π 4 A2 λ4